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Everyone. I've been trying to solve this task for an hour now and I got pretty nervous so i had to ask for help. The task is to find the parameter $m$ ($m$ belongs to the set of real numbers) for which the function $$y=(m+1)x^2 + (m-5)|x| +m-2$$ has the function minimum for $x=1$. I tried to use $x=1$ in the function formula and then use canonical form of quadratic function to get $$\alpha =\frac{m-5}{2(m+1)}=-1$$ and the for $x>0$, the $m$ is $1$ and for the $x<0$ the $m=-1$. But I am not sure if this solution is correct since when I draw the function it doesn't look as it should.

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To find the minimum such $m$, wouldn't you differentiate the function with respect to $m$, set the derivative to zero, and then solve for $m$? That should yield a linear equation, not a quadratic one. – MJD Mar 22 '12 at 21:00
Do you mean $\alpha =(m-5)/(2(m+1))=-1$ instead of $\alpha =(m-5)/2(m+1)=-1$? – Américo Tavares Mar 22 '12 at 21:12
yes, sorry .... – Transcendental Mar 22 '12 at 21:14
up vote 1 down vote accepted

Let $f(x)$ be our function. For positive $x$, we have $f'(x)=2(m+1)x+m-5$. If we have a local minimum at $x=1$, we must have $f'(1)=0$. This implies that $2(m+1)+m-5=0$, or equivalently $m=1$. The value of $f(1)$ is then $-3$.

The only thing we need to worry about is what happens for negative $x$. Our curve has obvious symmetry about the $y$-axis. So for $m=1$, an absolute minimum value of $-3$ is also reached when $x=-1$.

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Thanks. Helped me alot :) – Transcendental Mar 22 '12 at 22:02

Since you are interested around $x=1$ you can drop the absolute value sign, then take the derivative with respect to $x$ getting $y'=2(m+1)x+(m-5)$ Now you want to find the $m$ that makes this zero at $x=1$, so $0=2(m+1)+m-5$ or $m=1$. You can see the result at this Alpha page

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